Câu 1:
2: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)
=>\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=2^2=4\)
=>\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}=\frac{2}{xy}-\frac{1}{z^2}\)
=>\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{z^2}+\frac{2}{yz}+\frac{2}{xz}=0\)
=>\(\left(\frac{1}{x^2}+\frac{2}{xz}+\frac{1}{z^2}\right)+\left(\frac{1}{z^2}+\frac{2}{yz}+\frac{1}{y^2}\right)=0\)
=>\(\left(\frac{1}{x}+\frac{1}{z}\right)^2+\left(\frac{1}{z}+\frac{1}{y}\right)^2=0\)
=>x=-z và y=-z
=>x=y=-z
Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)
=>\(\frac{1}{x}+\frac{1}{x}-\frac{1}{x}=2\)
=>\(\frac{1}{x}=2\)
=>\(x=\frac12\)
=>\(x=y=\frac12;z=-\frac12\)
\(P=\left(x+2y+z\right)^{2024}=\left(\frac12+2\cdot\frac12+\frac{-1}{2}\right)^{2024}=1^{2024}=1\)
1: ĐKXĐ: x∉{0;1;-1}
a: \(A=\frac{x^2+x}{x^2-2x+1}:\left(\frac{x+1}{x}+\frac{1}{x-1}+\frac{2-x^2}{x^2-x}\right)\)
\(=\frac{x\left(x+1\right)}{\left(x-1\right)^2}:\frac{\left(x+1\right)\left(x-1\right)+x+2-x^2}{x\left(x-1\right)}\)
\(=\frac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\frac{x\left(x-1\right)}{x^2-1+x+2-x^2}=\frac{x^2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{x^2}{x-1}\)
b: Để A<0 thì \(\frac{x^2}{x-1}<0\)
=>x-1<0
=>x<1
Kết hợp ĐKXĐ, ta được: \(\begin{cases}x<1\\ x\notin\left\lbrace0;-1\right\rbrace\end{cases}\)
c: \(A=\frac{x^2}{x-1}=\frac{x^2-x+x-1+1}{x-1}=\left(x+1\right)+\frac{1}{x-1}=x-1+\frac{1}{x-1}+2\)
=>\(A\ge2\cdot\sqrt{\left(x-1\right)\cdot\frac{1}{x-1}}+2=2+2=4\forall x\) >1
Dấu '=' xảy ra khi x-1=1
=>x=2
